Question

Find the position vector of a point $R$ which divides the line joining two points $P$ and $Q$ whose position vectors are $(2\overrightarrow{a}+\overrightarrow{b})$ and $(\overrightarrow{a}-3\overrightarrow{b})$ externally in the ratio $1:2$. Also, show that $P$ is the mid point of the line segment $RQ$.

Answer 1

Let the position vector of $P$ be $\overrightarrow{OP}=2\overrightarrow{a}+\overrightarrow{b}$

Position vector of $Q$ be $\overrightarrow{OQ}=\overrightarrow{a}-3\overrightarrow{b}$

Now it is given the point R divides the line PQ externally in the ratio $1:2$

Therefore position vector of $R=\frac{m\overrightarrow{OQ}-n\overrightarrow{OP}}{m-n}$

$=\frac{\left(1(\overrightarrow{a}-3\overrightarrow{b})\right)-2(2\overrightarrow{a}+\overrightarrow{b})}{1-2}$

$=3\overrightarrow{a}+5\overrightarrow{b}$

Midpoint of $RQ=\frac{\overrightarrow{OQ}+\overrightarrow{OR}}{2}$

$=\frac{(\overrightarrow{a}-3\overrightarrow{b})+(3\overrightarrow{a}+5\overrightarrow{b})}{2}$

$=\frac{(4\overrightarrow{a}+2\overrightarrow{b})}{2}$

$=2\overrightarrow{a}+\overrightarrow{b}$

$=\overrightarrow{OP}$

Hence, $P$ is the mid point of segment $RQ$